Problem: Find $\dfrac{d}{dx}\left[\dfrac{e^{ 2x}}{2x}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^{ 2x}\left(2x-1\right)}{2x^2}$ (Choice B) B $\dfrac{e^{ 2x}\left(x-1\right)}{x}$ (Choice C) C $\dfrac{e^{ 2x}}{2}$ (Choice D) D $e^{ 2x}$
$\dfrac{e^{ 2x}}{2x}$ is a quotient of a composite function and another function. Let... $u(x)=e^{ x}$ $v(x)=2x$ $w(x)=2x$... then $\dfrac{e^{ 2x}}{2x}=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $\dfrac{d}{dx}\left[ \dfrac{e^{ 2x}}{2x}\right]$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=e^x$ $v'(x)=2$ $w'(x)=2$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{e^{ 2x}}(2)(2x)-{e^{ 2x}}\cdot 2}{(2x)^2} \\\\ &=\dfrac{2e^{ 2x}\left(2x-1\right)}{4x^2} \\\\ &=\dfrac{e^{ 2x}\left(2x-1\right)}{2x^2} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[ \dfrac{e^{ 2x}}{2x}\right]=\dfrac{e^{ 2x}\left(2x-1\right)}{2x^2}$.